JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 6)
If $${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$ (x > $$3 \over 4$$), then x is equal to :
$${{\sqrt {145} } \over {10}}$$
$${{\sqrt {145} } \over {11}}$$
$${{\sqrt {145} } \over {12}}$$
$${{\sqrt {146} } \over {12}}$$
Explanation
Given,
$${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$
$$ \Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$$
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]$$
$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}$$
$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}$$
$$ \Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}$$
$$ \Rightarrow 64 = 9\left( {16{x^2} - 9} \right)$$
$$ \Rightarrow 16{x^2} - 9 = {{64} \over 9}$$
$$ \Rightarrow 16{x^2} = {{64} \over 9} + 9$$
$$ \Rightarrow 16{x^2} = - {{145} \over 9}$$
$$ \Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}$$
$$ \Rightarrow x = \pm {{\sqrt {145} } \over {12}}$$
as given that $$x > {3 \over 4}$$
$$ \therefore $$ x $$ \ne $$ $$ - {{\sqrt {145} } \over {12}}$$
$$ \therefore $$ x $$ = {{\sqrt {145} } \over {12}}$$
$${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$
$$ \Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$$
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]$$
$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}$$
$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}$$
$$ \Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}$$
$$ \Rightarrow 64 = 9\left( {16{x^2} - 9} \right)$$
$$ \Rightarrow 16{x^2} - 9 = {{64} \over 9}$$
$$ \Rightarrow 16{x^2} = {{64} \over 9} + 9$$
$$ \Rightarrow 16{x^2} = - {{145} \over 9}$$
$$ \Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}$$
$$ \Rightarrow x = \pm {{\sqrt {145} } \over {12}}$$
as given that $$x > {3 \over 4}$$
$$ \therefore $$ x $$ \ne $$ $$ - {{\sqrt {145} } \over {12}}$$
$$ \therefore $$ x $$ = {{\sqrt {145} } \over {12}}$$
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